Weak boundary value problem with an inverse-square potential
= Weak boundary value problem with an inverse-square potential
On $H=\{v\in H^1(0,1):v(0)=0\}$, the form
$$
a(u,v)=\int_0^1u'v'+\int_0^1\frac{uv}{x^2}-\int_0^1u'v
$$
is bounded and coercive. The <Hardy inequality on an interval> controls its singular term, while the one-sided <Poincare inequality> controls the first-order term. Thus the <Lax-Milgram theorem> gives a unique weak solution for every functional $v\mapsto\int_0^1(f/x)(v/x)$ with $f/x\in L^2(0,1)$.